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Integration cos^2(∏/2cosθ)

by doey
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doey
#1
Dec3-12, 09:16 AM
P: 27
i facing a maths problem in integrating ∫ cos^2(∏/2cosθ) with limit from 0 to ∏/2,i was panic and struggled a long period of time in solving this,anyone can help me? pls give me the answer in detail tq !
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Millennial
#2
Dec3-12, 10:13 AM
P: 295
Use the integral identity [itex]\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx[/itex].
arildno
#3
Dec3-12, 10:32 AM
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Why would that help, Millenial?
You change the internal cos(theta) to a sin(theta)..

micromass
#4
Dec3-12, 10:40 AM
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Integration cos^2(∏/2cosθ)

Quote Quote by doey View Post
i facing a maths problem in integrating ∫ cos^2(∏/2cosθ) with limit from 0 to ∏/2,i was panic and struggled a long period of time in solving this,anyone can help me? pls give me the answer in detail tq !
First of all, do you mean

(1) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta[/itex]

or

(2) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2\cos \theta}) d\theta[/itex]

Also, which methods do you have at your disposal? Contour integration? Differentiation under the integral sign? Just normal calc II techniques?
doey
#5
Dec3-12, 11:02 AM
P: 27
Quote Quote by micromass View Post
First of all, do you mean

(1) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta[/itex]

or

(2) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2\cos \theta}) d\theta[/itex]

Also, which methods do you have at your disposal? Contour integration? Differentiation under the integral sign? Just normal calc II techniques?
(1) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta[/itex]
,i am asking this pls let me know the steps it takes
micromass
#6
Dec3-12, 11:21 AM
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Quote Quote by doey View Post
(1) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta[/itex]
,i am asking this pls let me know the steps it takes
Which course is this for? Do you know Bessel functions?? The solution requires this (at least that is what wolfram alpha says).
arildno
#7
Dec3-12, 11:33 AM
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Just a thought:
We may easily rewrite this equation into the identity:
[tex]\int_{0}^{\frac{\pi}{2}}\cos^{2}(\frac{\pi}{2}\cos\theta)d\theta+\int_{ 0}^{\frac{\pi}{2}}\sin^{2}( \frac{\pi}{2}\cos\theta)d\theta=\frac{\pi}{2}[/tex]
I feel dreadfully tempted to declare the two integrals to have the same value (the latter being merely a flipped version of the first), but temptation is not proof..
micromass
#8
Dec3-12, 11:39 AM
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Quote Quote by arildno View Post
Just a thought:
We may easily rewrite this equation into the identity:
[tex]\int_{0}^{\frac{\pi}{2}}\cos^{2}(\frac{\pi}{2}\cos\theta)d\theta+\int_{ 0}^{\frac{\pi}{2}}\sin^{2}( \frac{\pi}{2}\cos\theta)d\theta=\frac{\pi}{2}[/tex]
I feel dreadfully tempted to declare the two integrals to have the same value (the latter being merely a flipped version of the first), but temptation is not proof..
Hmmm, looking at the graph doesn't really convince me that the integrals are equal

Anyway, wolfram alpha gives us

[itex]\int_0^{\pi/2} \cos^2(\frac{\pi}{2} \cos(x))dx = \frac{\pi}{4}(1+J_0(\pi))[/itex]

so I doubt the integral will be solvable with methods like these.
Mute
#9
Dec3-12, 04:32 PM
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To evaluate the integral one will have to use the identities

$$\cos t = \frac{1}{2}(e^{it}+e^{-it})$$
(or just ##\cos t = \mbox{Re}[\exp(it)]##)

and

$$e^{iz\cos\theta} = \sum_{n=-\infty}^\infty i^n J_n(z)e^{in\theta}.$$

I guess the trig identity

$$\cos^2 t = \frac{1}{2}(1+\cos(2t))$$
also helps.


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